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If u, vand w (all positive) are the $\mathrm{p}^{\mathrm{th}}, \mathrm{q}^{\text {th }}$ and $\mathrm{r}^{\text {th }}$ terms of a GP, the determinant of the matrix
$\left(\begin{array}{l}\ln \mathrm{u} \mathrm{p} 1 \\ \ln \mathrm{v} \mathrm{q} 1 \\ \ln \mathrm{w} \mathrm{r} 1\end{array}\right)$ is
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$\left(\begin{array}{l}\ln \mathrm{u} \mathrm{p} 1 \\ \ln \mathrm{v} \mathrm{q} 1 \\ \ln \mathrm{w} \mathrm{r} 1\end{array}\right)$ is
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Let $u, v$ and $w$ are $\mathrm{p}^{\text {th }}, q^{\text {th }}$ and $\mathrm{r}^{\text {th }}$ term of the G.P. with first term ' $a$ ' and common ratio ' $d$ '.
then, $u=a .(d)^{p-1} \Rightarrow \operatorname{In} u=\operatorname{In}(a)+(p-1) \operatorname{In}(d)$
$v=a \cdot(d)^{q-1} \Rightarrow \operatorname{In} v=\operatorname{In}(a)+(q-1) \operatorname{In}(d)$
$w=a .(d)^{r-1} \Rightarrow \operatorname{In} w=\operatorname{In}(a)+(r-1) \operatorname{In}(d)$
Now, $\operatorname{In} u-\operatorname{In} v=(p-q) \operatorname{In}(d)$
In $u-\operatorname{In} w=(p-r) \operatorname{In}(d)$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} v & q & 1 \\ \operatorname{In} w & r & 1\end{array}\right|$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} u-\operatorname{In} v & (p-q) & 0 \\ \operatorname{In} u-\operatorname{In} w & (p-r) & 0\end{array}\right|$
\{Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\right\}$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ (p-q) \operatorname{In}(d) & (p-q) & 0 \\ (p-r) \operatorname{In}(d) & (p-r) & 0\end{array}\right|$
$=(p-q)(p-r)[\operatorname{In}(d)-\operatorname{In}(d)]=0$
then, $u=a .(d)^{p-1} \Rightarrow \operatorname{In} u=\operatorname{In}(a)+(p-1) \operatorname{In}(d)$
$v=a \cdot(d)^{q-1} \Rightarrow \operatorname{In} v=\operatorname{In}(a)+(q-1) \operatorname{In}(d)$
$w=a .(d)^{r-1} \Rightarrow \operatorname{In} w=\operatorname{In}(a)+(r-1) \operatorname{In}(d)$
Now, $\operatorname{In} u-\operatorname{In} v=(p-q) \operatorname{In}(d)$
In $u-\operatorname{In} w=(p-r) \operatorname{In}(d)$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} v & q & 1 \\ \operatorname{In} w & r & 1\end{array}\right|$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} u-\operatorname{In} v & (p-q) & 0 \\ \operatorname{In} u-\operatorname{In} w & (p-r) & 0\end{array}\right|$
\{Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\right\}$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ (p-q) \operatorname{In}(d) & (p-q) & 0 \\ (p-r) \operatorname{In}(d) & (p-r) & 0\end{array}\right|$
$=(p-q)(p-r)[\operatorname{In}(d)-\operatorname{In}(d)]=0$
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